Rectangular hyperbola xy = c²: Exercises — Maths deep dive

The blueprint above is our reference board: the two axes are the asymptotes, the amber dots are two sample points $(c t_{1}, c / t_{1})$ and $(c t_{2}, c / t_{2})$ , and the cyan line through them is a chord. Keep it in view.

Level 1 — Recognition

Recall Solution

WHAT: Check the defining equation. WHY: "on the curve" literally means the coordinates satisfy $x y = c^{2}$ . Here $c^{2} = 16 \Rightarrow c = 4$ . Test: $x \cdot y = 8 \cdot 2 = 16 = c^{2}$ ✓ — yes, it lies on the curve. Parameter: $x = c t \Rightarrow 8 = 4 t \Rightarrow t = 2$ . Confirm $y = c / t = 4/2 = 2$ ✓. Answer: on the curve, with $t = 2$ .

Recall Solution

WHAT/WHY: Pure recall of the "Key facts" box; recognising standard structure. $c^{2} = 25 \Rightarrow c = 5$ . Vertices $(c, c), (- c, - c) = (5, 5)$ and $(- 5, - 5)$ , sitting on the transverse axis $y = x$ . Eccentricity $e = 2$ (true for every rectangular hyperbola). Asymptotes: $x = 0$ (the $y$ -axis) and $y = 0$ (the $x$ -axis) — perpendicular, which is exactly what "rectangular" means.

Level 2 — Application

Recall Solution

WHAT: Plug into the derived tangent form. WHY: never re-derive; the formula $\frac{x}{t} + t y = 2 c$ already encodes the implicit differentiation. $c^{2} = 9 \Rightarrow c = 3$ . With $t = 3$ : $\frac{x}{3} + 3 y = 6 ⟹ x + 9 y = 18.$ Axis intercepts (WHY: the intercepts are needed for the area property later):

Cut the $x$ -axis ( $y = 0$ ): $x = 18$ , point $(18, 0)$ .
Cut the $y$ -axis ( $x = 0$ ): $9 y = 18 \Rightarrow y = 2$ , point $(0, 2)$ . Answer: tangent $x + 9 y = 18$ ; it meets the axes at $(18, 0)$ and $(0, 2)$ .

Recall Solution

WHAT: Use the normal form $x t^{3} - y t = c (t^{4} - 1)$ . WHY: the normal is perpendicular to the tangent, so its slope is the negative reciprocal of $- 1/ t^{2}$ , giving $+ t^{2}$ ; the boxed formula bakes this in. $c^{2} = 4 \Rightarrow c = 2$ , $t = 1$ : $x (1) - y (1) = 2 (1 - 1) = 0 ⟹ x - y = 0, i.e. y = x .$ What it looks like: at $t = 1$ the point is $(c, c) = (2, 2)$ , a vertex on the transverse axis $y = x$ ; the normal there is the transverse axis. Slope $+ 1 = t^{2}$ ✓.

Recall Solution

WHAT: Apply $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ . WHY: the parametric chord formula spares us computing a slope from two messy fractions. $c^{2} = 6 \Rightarrow c = 6$ . Then $t_{1} t_{2} = 3$ , $t_{1} + t_{2} = 4$ : $x + 3 y = 46 .$ Self-check: the endpoints are $(6, 6)$ and $(36, 6 /3)$ . First: $6 + 36 = 46$ ✓. Second: $36 + 3 (6 /3) = 36 + 6 = 46$ ✓.

Level 3 — Analysis

Recall Solution

WHAT: Use the derived meeting point $(\frac{2 c t _{1} t _{2}}{t _{1} + t _{2}}, \frac{2 c}{t _{1} + t _{2}})$ . WHY: solving two tangent lines simultaneously every time is slow; the symmetric formula is the shortcut. $c^{2} = 8 \Rightarrow c = 22$ . Here $t_{1} t_{2} = 4$ , $t_{1} + t_{2} = 5$ . $x = \frac{2 ( 2 2 ) ( 4 )}{5} = \frac{16 2}{5}, y = \frac{2 ( 2 2 )}{5} = \frac{4 2}{5} .$ Answer: $(\frac{16 2}{5}, \frac{4 2}{5})$ . Sanity: the $x$ -coordinate is $t_{1} t_{2} = 4$ times the $y$ -coordinate ( $162 /5 = 4 \cdot 42 /5$ ✓), matching $x = t_{1} t_{2} y$ found in the parent derivation.

Recall Solution

WHAT: The chord endpoints are $(c t_{1}, c / t_{1}), (c t_{2}, c / t_{2})$ ; the midpoint averages them. WHY: relating $t_{1} + t_{2}, t_{1} t_{2}$ to $h, k$ turns a geometric condition into two algebraic handles. Midpoint: $h = \frac{c t _{1} + c t _{2}}{2} = \frac{c ( t _{1} + t _{2} )}{2} \Rightarrow t_{1} + t_{2} = \frac{2 h}{c},$ $k = \frac{1}{2} (\frac{c}{t _{1}} + \frac{c}{t _{2}}) = \frac{c ( t _{1} + t _{2} )}{2 t _{1} t _{2}} = \frac{h}{t _{1} t _{2}} \Rightarrow t_{1} t_{2} = \frac{h}{k} .$ Now plug into the chord form $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ : $x + \frac{h}{k} y = c \cdot \frac{2 h}{c} = 2 h .$ Divide by $h$ : $\frac{x}{h} + \frac{y}{hk / h} \dots$ — cleanly, dividing $x + \frac{h}{k} y = 2 h$ by $h$ gives $\frac{x}{h} + \frac{y}{k} = 2. ■$ Result: $t_{1} + t_{2} = \frac{2 h}{c}$ , $t_{1} t_{2} = \frac{h}{k}$ , and the chord bisected at $(h, k)$ is $\frac{x}{h} + \frac{y}{k} = 2$ .

The figure shows the chord (cyan) with midpoint $(h, k)$ (amber): both endpoints are equally far from $(h, k)$ along the line — the averaging of the two parametric points is exactly what the algebra above computed.

Level 4 — Synthesis

Recall Solution

WHAT: find the two axis-intercepts of the tangent, then the triangle area $\frac{1}{2} (base) (height)$ . WHY: $O$ is the right-angle corner (axes are perpendicular), so the intercept lengths are the two legs — area is simply half their product. Tangent: $\frac{x}{t} + t y = 2 c$ .

$y = 0$ (x-axis): $\frac{x}{t} = 2 c \Rightarrow x = 2 c t$ , so $P = (2 c t, 0)$ .
$x = 0$ (y-axis): $t y = 2 c \Rightarrow y = \frac{2 c}{t}$ , so $Q = (0, \frac{2 c}{t})$ . Legs: $∣ O P ∣ = ∣2 c t ∣$ , $∣ O Q ∣ = \frac{2 c}{t}$ . Since the axes meet at $9 0^{\circ}$ , $Area = \frac{1}{2} ∣ O P ∣ ∣ O Q ∣ = \frac{1}{2} ∣2 c t ∣ \cdot \frac{2 c}{t} = \frac{1}{2} \cdot 4 c^{2} = 2 c^{2} .$ Result: area $= 2 c^{2}$ , the $t$ cancels — constant. $■$

What it looks like: as $t$ grows, $P$ slides far out along one axis while $Q$ creeps toward $O$ on the other; the triangle stretches thin but its area is pinned at $2 c^{2}$ . Compare with the parent's rectangle area $c^{2}$ : the tangent triangle is exactly twice the corner rectangle.

Recall Solution

WHAT: confirm the two contact points $(c t_{1}, c / t_{1})$ and $(c t_{2}, c / t_{2})$ satisfy the chord equation. WHY: the "chord of contact" from an external point $M$ is precisely the line through the two touch-points; verifying membership proves the geometry. Chord: $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ .

Contact point $t_{1}$ : $c t_{1} + t_{1} t_{2} \cdot \frac{c}{t _{1}} = c t_{1} + c t_{2} = c (t_{1} + t_{2})$ ✓.
Contact point $t_{2}$ : $c t_{2} + t_{1} t_{2} \cdot \frac{c}{t _{2}} = c t_{2} + c t_{1} = c (t_{1} + t_{2})$ ✓. Both lie on the chord, so the chord is the chord of contact from $M$ . Combined with L3·Q1 ( $M$ satisfies $x = t_{1} t_{2} y$ and $y = \frac{2 c}{t _{1} + t _{2}}$ ), $M$ is the pole. $■$

Level 5 — Mastery

Recall Solution

WHAT: the normal is a line; where it re-hits the curve is a chord from $t_{1}$ to $t_{2}$ , so it must coincide with the chord equation. WHY: a line through two curve points is the chord $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ ; the same line written as the normal gives a second description. Matching the two forces a relation on $t_{1}, t_{2}$ . Normal at $t_{1}$ : $x t_{1}^{3} - y t_{1} = c (t_{1}^{4} - 1)$ . Divide by $t_{1}$ (allowed, $t_{1} \neq = 0$ ): $x t_{1}^{2} - y = \frac{c ( t _{1}^{4} - 1 )}{t _{1}} ⟹ x t_{1}^{2} - y = c (t_{1}^{3} - \frac{1}{t _{1}}) .$ Chord $t_{1} t_{2}$ : $x + t_{1} t_{2} y = c (t_{1} + t_{2})$ . Rewrite as $\frac{x}{- t _{1} t _{2}} \dots$ — instead compare slopes and one point. Slope of the normal is $+ t_{1}^{2}$ . Slope of the chord is $- \frac{1}{t _{1} t _{2}}$ (parent note). Equate: $t_{1}^{2} = - \frac{1}{t _{1} t _{2}} ⟹ t_{1}^{3} t_{2} = - 1. ■$ What it looks like: since $t_{2} = - 1/ t_{1}^{3}$ has the opposite sign to $t_{1}$ , the normal from a point on one branch strikes the other branch (or far along the same one) — the curve's two arms are stitched by every normal.

Recall Solution

WHAT: use $t_{1} + t_{2} = \frac{2 h}{c}$ , $t_{1} t_{2} = \frac{h}{k}$ (from L3·Q2) plus $t_{1}^{3} t_{2} = - 1$ . WHY: three equations in $t_{1}, t_{2}, h, k$ — eliminate $t_{1}, t_{2}$ to leave a curve in $(h, k)$ , i.e. the locus. From $t_{1} t_{2} = \frac{h}{k}$ and $t_{1}^{3} t_{2} = - 1$ , divide: $\frac{t _{1}^{3} t _{2}}{t _{1} t _{2}} = t_{1}^{2} = \frac{- 1}{h / k} = - \frac{k}{h} ⟹ t_{1}^{2} = - \frac{k}{h} .$ (This forces $h, k$ to have opposite signs, consistent with the normal crossing branches.) Also $t_{1} t_{2} = \frac{h}{k}$ gives $t_{2} = \frac{h}{k t _{1}}$ . Use $t_{1} + t_{2} = \frac{2 h}{c}$ : $t_{1} + \frac{h}{k t _{1}} = \frac{2 h}{c} .$ Multiply by $t_{1}$ and substitute $t_{1}^{2} = - \frac{k}{h}$ : $t_{1}^{2} + \frac{h}{k} = \frac{2 h}{c} t_{1} \Rightarrow - \frac{k}{h} + \frac{h}{k} = \frac{2 h}{c} t_{1} \Rightarrow t_{1} = \frac{c}{2 h} (\frac{h}{k} - \frac{k}{h}) = \frac{c ( h ^{2} - k ^{2} )}{2 hk h} .$ So $t_{1} = \frac{c ( h ^{2} - k ^{2} )}{2 h ^{2} k}$ . Square and set equal to $t_{1}^{2} = - \frac{k}{h}$ : $\frac{c ^{2} ( h ^{2} - k ^{2} ) ^{2}}{4 h ^{4} k ^{2}} = - \frac{k}{h} \Rightarrow c^{2} (h^{2} - k^{2})^{2} = - 4 h^{3} k^{3} .$ Replacing $(h, k) \to (x, y)$ , the locus is $c^{2} (x^{2} - y^{2})^{2} + 4 x^{3} y^{3} = 0.$ Answer / check of sign: since $x, y$ have opposite signs on such chords, $x^{3} y^{3} < 0$ , so $- 4 x^{3} y^{3} > 0$ balances the non-negative left term — the equation is consistent. $■$

Answer Key (quick reference)

Q	Answer
L1·Q1	On curve, $t = 2$
L1·Q2	$c = 5$ ; vertices $(5, 5), (- 5, - 5)$ ; $e = 2$ ; asymptotes $x = 0, y = 0$
L2·Q1	$x + 9 y = 18$ ; intercepts $(18, 0), (0, 2)$
L2·Q2	$y = x$
L2·Q3	$x + 3 y = 46$
L3·Q1	$(\frac{16 2}{5}, \frac{4 2}{5})$
L3·Q2	$\frac{x}{h} + \frac{y}{k} = 2$ ; $t_{1} + t_{2} = \frac{2 h}{c}$ , $t_{1} t_{2} = \frac{h}{k}$
L4·Q1	Area $= 2 c^{2}$
L5·Q1	$t_{1}^{3} t_{2} = - 1$
L5·Q2	$c^{2} (x^{2} - y^{2})^{2} + 4 x^{3} y^{3} = 0$

Connections

Parent: Rectangular hyperbola $xy=c^2$ — all formulas used here are derived there.
Tangent and Normal to conics — the tangent/normal machinery driving L2–L5.
Asymptotes of a hyperbola — the axes that L4's triangle rests on.
Hyperbola standard form — where these parametric ideas first appear.
Reciprocal function y=1/x — the plotted shape behind every figure.